Are you talking about for all values of n? Or specific values of n? Obviously if it's the last one, every function fits n = 1....

-Dan

Apr 26th 2008, 02:46 PM

mmzaj

i'm talking about all positive integers .
my bad (Wink)
$\displaystyle
\phi(t) = 0 \
$
is rather a trivial solution that i'm not interested in
i know that there will be a family of solutions , i'm interested in this family .

thanks , but there is something wrong , 'cause $\displaystyle \phi(t) $ should be independent of n . the equality is required to be true for any positive integer ... so - obviously - $\displaystyle \phi(t) $ should be independent of n .

here is what i did :

assume $\displaystyle \phi(t) $ is smooth and analytic at $\displaystyle t=t_0 $ , then it can be expanded in terms of a power series . now setting the correct relations on both RH and LH parts , and integrating over T , we end up with something like this :

now the program is to solve for $\displaystyle b_r $ in general .. so , is this doable ?

Apr 27th 2008, 07:20 PM

mr fantastic

Quote:

Originally Posted by mmzaj

thanks , but there is something wrong , 'cause $\displaystyle \phi(t) $ should be independent of n . the equality is required to be true for any positive integer ... so - obviously - $\displaystyle \phi(t) $ should be independent of n .
[snip]

I disagree. A solution containing n is perfectly fine ...... I think the only solution you'll find that's independent of n will be the trivial solution $\displaystyle \phi(t) = 0$ ....